The limit of difference between $E\ln (1+X)$ and $\ln(E(1+X))$

Question

Define a random variable $X=v^2$ where $v$ is Gaussian variable with mean $m$ and variance $\sigma^2$. I am interested in whether $$\lim_{m \rightarrow \infty} (E\ln(1+X) - \ln(E(1+X))) \rightarrow0 $$

where the variance $\sigma^2$ still remains a constant.

The following is the result of difference between $E\ln(1+X)$ and $\ln(E(1+X))$ as $m$ increases exponentially.

We can see the difference approaches to 0 as the increase of $m$. So I guess $\ln(E(1+X))$ is the upper bound of $E(\ln (1+X))$. Is it true and how to prove or disprove it?

Thank you!

Answer 1

Henry's example shows why you're seeing $\ln(E(1+X))$ as an upper bound.

Lets try to prove that this difference converges to 0

Let $Y_m:= \ln(1+X_m)$

Note that $E[1+X_m]=m^2+\sigma^2+1$ and $E[Y_m]\geq 0$ by its definition.

By Markov's Inequality:

$$P(Y_m>a)\leq \frac{E[Y_m]}{a}\implies E[Y_m]\geq aP(Y_m>a)$$

However, by Jensen's Inequality:

$$\ln(1+m^2+\sigma^2)\geq E[Y_m]$$

Combining these gives:

$$aP(Y_m>a)\leq E[Y_m]\leq\ln(1+m^2+\sigma^2)$$

Now:

$$P(Y_m>a)=P(X_m<e^a-1)=P(v_m>\sqrt{e^a-1})+P(v_m<-\sqrt{e^a-1})$$

Standardizing $v_m$ we get:

$$P(Y_m>a)=1-\Phi\left(\frac{\sqrt{e^a-1}-m}{\sigma}\right)+\Phi\left(\frac{-\sqrt{e^a-1}-m}{\sigma}\right)$$

Thus:

$$a\left[1-\Phi\left(\frac{\sqrt{e^a-1}-m}{\sigma}\right)+\Phi\left(\frac{-\sqrt{e^a-1}-m}{\sigma}\right)\right]\leq E[Y_m]\leq \ln(1+m^2+\sigma^2),\;\forall a>0$$

Lets set $a=k\ln(1+m^2+\sigma^2),k\in(0,1)$:

$$k\ln(1+m^2+\sigma^2)\left[1-\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)+\Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)\right]\leq E[Y_m]\leq \ln(1+m^2+\sigma^2)$$

Taking the limit of the LHS:

$$\lim_{m\to \infty} k\ln(1+m^2+\sigma^2)\left[1-\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)+\Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)\right] = k\ln(1+m^2+\sigma^2)$$

Since $\forall k \in (0,1): (m^2+\sigma^2)^{0.5k}-m=O(m)$; therefore,

$$ k\ln(1+m^2+\sigma^2)\leq E[Y_m]\leq \ln(1+m^2+\sigma^2)$$

Maximizing the lower bound by letting $k\to 1$ gives:

$$ \ln(1+m^2+\sigma^2)\leq E[Y_m]\leq \ln(1+m^2+\sigma^2) \implies \lim_{m\to \infty} E[Y_m]=\ln(1+m^2+\sigma^2)$$

$\square$

Answer 2

This is essentially an example of Jensen's inequality as the logarithm is a concave function $f$ with the property $$f(E[X]) \ge E[f(X)]$$ resulting from the concavity of $f$.

The direction of the inequality would reverse for a convex function.

Answer 3

Let $Y=1+X$, By Jensen's inequality, we can have $$E(Y)<\ln(E(1+X))=\ln(1+m^2+\sigma^2)$$ Since $Y=\ln (1+v^2)>0$, apply Markov inequality, for any $a>0$, $$P(Y>a)\leq \frac{EY}{a}$$ $$EY\geq aP(Y>a)$$ So we have $$aP(Y>a)\leq EY \leq \ln(1+m^2+\sigma^2)$$ Set $a=k \ln(1+m^2+\sigma^2)$, where $0<k<1$. $$P(Y>a)=1-\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)+\Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)$$ So $$aP(Y>a)=k\ln(1+m^2+\sigma^2)(1-\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)+\Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right))$$

Next we show that $$D=\lim_{m \rightarrow \infty} k\ln(1+m^2+\sigma^2)\left[\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)-\Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)\right]=0$$

Firstly because $\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right) \geq \Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)$ So $D \geq 0$.

And $D \leq \lim_{m \rightarrow \infty} \ln(1+m^2+\sigma^2)\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)$

Let $M=(m^2+\sigma^2)^{0.5k}-m$, note that because $k<1$, so the order of $M$ is the same as $m$, and $M \rightarrow -\infty$ as $m \rightarrow \infty$

$$ \begin{aligned} D &\leq \frac{1}{\sqrt{2\pi}} \lim_{m \rightarrow \infty} \ln(1+m^2+\sigma^2) \int_{-\infty}^{M} e^\frac{-x^2}{2} \, dx \\ &\leq \frac{1}{\sqrt{2\pi}} \lim_{m \rightarrow \infty} \ln(1+m^2+\sigma^2) \int_{-M}^{\infty} e^{-x} \, dx \\ &=\frac{1}{\sqrt{2\pi}} \lim_{m \rightarrow \infty} \ln(1+m^2+\sigma^2)e^M=0 \end{aligned} $$

Then we have $$\lim_{m \rightarrow \infty} aP(Y>a)=k\ln(1+m^2+\sigma^2)$$ Because the choice of $k$ is arbitrary except $0<k<1$, let's choose $k$ approach to 1. Then $$\lim_{k \rightarrow 1} \lim_{m \rightarrow \infty} aP(Y>a)=\ln(1+m^2+\sigma^2)$$

$$\ln (1+m^2+\sigma^2)\leq \lim_{m \rightarrow \infty} E(Y) \leq \ln (1+m^2+\sigma^2)$$

So $$\lim_{m \rightarrow \infty} E(Y) = \ln(1+m^2+\sigma^2)$$